Explanation Given: An undirected graph is with n vertices. Proof: Let C be any cycle in G and let the edge e = ( v , w ) be the most important edge in C . The edge costs are all distinct. Prove that e does not belong to any minimum spanning tree of G . Let T be a spanning tree that covers e . Remove e partitions the vertices into two components: A part S that contain v , and V − S contain w . The edges of C other than e from path P with one end at v and other at w

Discrete Mathematics And Its Applications

8th Edition

ISBN: 9781260091991

Author: NA

Publisher: Mc graw hill

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Chapter 11.5, Problem 33E

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If G is a weighted graph with distinct edge weights, then for every simple circuit of G ¸ the edge of maximum weight in this circuit does not belong to any minimum spanning tree of G.

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Chapter 11.5, Problem 32E

Chapter 11.5, Problem 34E

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Chapter 11 Solutions

Discrete Mathematics And Its Applications

Ch. 11.1 - Prob. 1E Ch. 11.1 - Vhich of these graphs are trees?Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

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If G is a weighted graph with distinct edge weights, then for every simple circuit of G ¸ the edge of maximum weight in this circuit does not belong to any minimum spanning tree of G .

Solution Summary: The author explains that if G is a weighted graph with distinct edge weights, the edge of maximum weight cannot be involved in the minimum spanning tree of the graph.

To show:

If G is a weighted graph with distinct edge weights, then for every simple circuit of G ¸ the edge of maximum weight in this circuit does not belong to any minimum spanning tree of G.

Want to see the full answer?

Chapter 11 Solutions

If G is a weighted graph with distinct edge weights, then for every simple circuit of G ¸ the edge of maximum weight in this circuit does not belong to any minimum spanning tree of G .

Solution Summary: The author explains that if G is a weighted graph with distinct edge weights, the edge of maximum weight cannot be involved in the minimum spanning tree of the graph.

To show: If G is a weighted graph with distinct edge weights, then for every simple circuit of G ¸ the edge of maximum weight in this circuit does not belong to any minimum spanning tree of G.

Want to see the full answer?

Chapter 11 Solutions

To show:

If G is a weighted graph with distinct edge weights, then for every simple circuit of G ¸ the edge of maximum weight in this circuit does not belong to any minimum spanning tree of G.